Mathematics Standard • Year 11 • Module 1 • Lesson 1

Algebraic Language, Variables and Substitution

Build fluency reading algebraic notation, substituting into formulas, and handling negative values with brackets.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Write the meaning of each piece of algebraic notation.

3x means ____________ x² means ____________ x/2 means ____________

Q1.2 Circle the equations and underline the expressions.

(a) 2x + 5 (b) C = 8 + 1.50k (c) lw (d) A = lw (e) 7m − 3

Q1.3 Identify the coefficient and the constant in the expression 4x + 9. Coefficient = ____________ Constant = ____________

Stuck? Revisit lesson § Key Terms — coefficient is the number multiplying the variable.

2. Worked example — substitute into a cost formula

Follow each line of working. Every step has a reason on the right.

Problem. A delivery company uses C = 8 + 1.50k, where C is the cost in dollars and k is the distance in kilometres. Find the cost for an 18 km delivery.

Step 1 — Identify the value to substitute.

k = 18

Reason: the question gives us the distance in km, which matches the variable k.

Step 2 — Substitute, using brackets around the number.

C = 8 + 1.50(18)

Reason: brackets make the multiplication explicit and prevent slips when the number is negative or has a decimal.

Step 3 — Apply order of operations (multiply before adding).

C = 8 + 27 = 35

Reason: 1.50 × 18 = 27, then add the 8 booking fee.

Conclusion. The delivery costs $35.00.

3. Faded example — fill in the missing steps

Use P = 2l + 2w to find the perimeter of a rectangle with l = 12 m and w = 5 m. Fill in each blank. 4 marks

Step 1 — Identify values:

l = ____________ w = ____________

Step 2 — Substitute, using brackets:

P = 2(____) + 2(____)

Step 3 — Calculate each term:

P = ____________ + ____________

Step 4 — Add and add the units:

P = ____________ m

Conclusion sentence. The perimeter is ____________ m.

Stuck? Revisit lesson § Worked Example 2 — Substitute into a perimeter formula.

4. Graduated practice — substitution

Show your working in the space below each part. Always write a units sentence at the end.

Foundation — single substitution, clean numbers (4 questions)

Q	Problem	Answer
4.1 1	Evaluate d = st when s = 60 km/h and t = 2 h.
4.2 1	Evaluate A = lw when l = 10 cm and w = 7 cm.
4.3 1	Evaluate C = 8 + 1.50k when k = 6.
4.4 1	Evaluate 3x + 4 when x = 5.

Standard — typical HSC difficulty (6 questions)

Show at least one line of substitution and label the final answer with units where relevant.

4.5 Use C = 12 + 2.40m to find the cost of a courier trip of m = 15 km. 2 marks

4.6 A rectangle has l = 11.5 cm and w = 8 cm. Use A = lw to find its area. 2 marks

4.7 Evaluate P = 2l + 2w when l = 8.5 m and w = 4.2 m. 2 marks

4.8 Evaluate 2x + 5 when x = −3. Use brackets in your substitution. 2 marks

4.9 Evaluate x² when x = −4. Show the bracketed substitution. 2 marks

4.10 Using d = st, find d when s = 95 km/h and t = 2.5 h. Write the answer in a units sentence. 2 marks

Extension — two-variable substitution and meaning (2 questions)

4.11 Evaluate x² + 3x when x = −4. Show every bracketed step. 3 marks

4.12 A taxi uses T = 4.20 + 2.30k for a trip of k kilometres. (a) Find T when k = 7.5. (b) Explain in one sentence what the 4.20 represents in the real-life context. 3 marks

Stuck on 4.11? Substitute x = −4 as (−4) into both terms, then square first, multiply second, then add.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I wrote d = 60 × 2 = 120 km and gave the answer with units.

For 4.3 I multiplied before adding (1.50 × 6 = 9, then 8 + 9 = 17).

For 4.4 I substituted x = 5 to get 3(5) + 4 = 19, not 35 (avoiding 3 + 5 = 8 then × 4).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Notation meanings

3x means 3 × x. x² means x × x. x/2 means x divided by 2.

Q1.2 — Equation vs expression

Equations (contain =): (b) C = 8 + 1.50k, (d) A = lw. Expressions: (a) 2x + 5, (c) lw, (e) 7m − 3.

Q1.3 — Coefficient and constant

Coefficient = 4 (the number multiplying x). Constant = 9 (the value that does not change).

Q3 — Faded perimeter example

Step 1: l = 12, w = 5.
Step 2: P = 2(12) + 2(5).
Step 3: P = 24 + 10.
Step 4: P = 34 m.
Conclusion: The perimeter is 34 m.

Q4.1 — d = 60 × 2

d = 60 × 2 = 120 km.

Q4.2 — A = 10 × 7

A = 10 × 7 = 70 cm².

Q4.3 — C = 8 + 1.50(6)

C = 8 + 9 = $17.00.

Q4.4 — 3x + 4 with x = 5

3(5) + 4 = 15 + 4 = 19.

Q4.5 — Courier C = 12 + 2.40(15)

C = 12 + 36 = $48.00 for the 15 km trip.

Q4.6 — A = 11.5 × 8

A = 11.5 × 8 = 92 cm².

Q4.7 — P = 2(8.5) + 2(4.2)

P = 17 + 8.4 = 25.4 m.

Q4.8 — 2x + 5 with x = −3

2(−3) + 5 = −6 + 5 = −1. (Bracketing the −3 prevents the slip "2−3+5 = 4".)

Q4.9 — x² with x = −4

(−4)² = (−4) × (−4) = 16. (Writing −4² without brackets would give −16 — wrong, because the square would only apply to the 4.)

Q4.10 — d = 95 × 2.5

d = 95 × 2.5 = 237.5. The distance travelled is 237.5 km.

Q4.11 — x² + 3x with x = −4

(−4)² + 3(−4) = 16 + (−12) = 16 − 12 = 4.

Q4.12 — Taxi formula T = 4.20 + 2.30k

(a) T = 4.20 + 2.30(7.5) = 4.20 + 17.25 = $21.45.
(b) The 4.20 is the flagfall (fixed booking/start fee) — the cost before any kilometres are travelled.